Coding Interview Questions

Alvin P. - "SELECT e1.empid AS manageremployee_id, e1.empname AS managername, COUNT(e2.empid) AS numberofdirectreports FROM employees AS e1 INNER JOIN employees AS e2 ON e2.managerid = e1.empid GROUP BY e1.emp_id HAVING COUNT(e2.emp_id) >= 2 ORDER BY numberofdirectreports DESC, managername ASC `"See full answer

Jonathan michael J. - "add two strings `"See full answer

Prachi G. - "The solution produces the same result as the 'prescribed solution' yet it does not get accepted In the test results section transcript['year'] = transcript['year'].astype(str) df = pd.pivottable(data = transcript, index = 'studentid', columns = 'year', values = 'yearlygpa', aggfunc = 'mean').resetindex() df = df[(df['2021'] < df['2022']) & (df['2022'] < df['2023'])] df['average_gpa'] = df[['2021', '2022', '2023']].mean(axis=1).round(2) return df "See full answer

Ina K. - "HashMap supports insert, search, delete and retrieve in O(1). It stores data as key value pairs."See full answer

Gowtami K. - "select a.playername as player1, b.playername as player2, abs(a.level-b.level) as level_disparity from players a cross join players b on a.playername < b.playername where abs(a.level-b.level) <=5 order by level_disparity `"See full answer

Derrick M. - "SELECT COUNT(DISTINCT o.customerid) AS customers, d.departmentname FROM orders o INNER JOIN departments d ON d.departmentid = o.departmentid WHERE d.departmentname IN ('Electronics','Fashion') AND o.orderdate BETWEEN '2022-01-01' AND '2022-12-31' GROUP BY d.department_name; `"See full answer

Divyani .. - "Did the code in Python"See full answer

The_ A. - "The best and average of both the algorithms is same which is O(nlog(n)), but the worst time complexity of QuickSort is O(n^2){a case where all the elements are sorted in opposite to the fashion/order you want} while the worst TC for merge sort remains the same O(nlog(n)). and The SC for QS=O(logn) and MS=O(n)."See full answer

Rg - "Used prefix & postfix sums, resetting the product when 0 was found"See full answer

Vince S. - "func isMatch(text: String, pattern: String) -> Bool { // Convert strings to arrays for easier indexing let s = Array(text.characters) let p = Array(pattern.characters) guard !s.isEmpty && !p.isEmpty else { return true } // Create DP table: dpi represents if s[0...i-1] matches p[0...j-1] var dp = Array(repeating: Array(repeating: false, count: p.count + 1), count: s.count + 1) // Empty pattern matches empty string dp[0]["See full answer

Gabriella F. - "Table user is empy....... Problem with this problem "See full answer

Flora K. - "USING RECURSION"See full answer

Nicholas S. - "Definitely nice to think of this without memorization, but there is a well known algorithm for this problem, which is the Levenshtein Distance. Lev(a,b) = len(a) if len(b) == 0 = len(b) if len(a) == 0 = lev(a[1:], b[1:] if a[0] == b[0] = 1 + min (lev(a, b[1:]), lev(a[1:], b), lev(a[1:], b[1:])) https://en.wikipedia.org/wiki/Levenshtein_distance I'm sure some optimizations could be made with heuristic."See full answer

Shivani N. - "did well but messed up dequeue"See full answer

Nishigandha B. - "na"See full answer